Tính đạo hàm của hàm số 1/2tanx -2sinx+cotx 26/09/2021 Bởi Charlie Tính đạo hàm của hàm số 1/2tanx -2sinx+cotx
$y’=\dfrac{(2\tan x-2\sin x+\cot x)’}{(2\tan x-2\sin x+\cot x)^2}$ $=\dfrac{2(\tan x)’-2(\sin x)’+(\cot x)’}{(2\tan x-2\sin x+\cot x)^2}$ $=\dfrac{\dfrac{2}{\cos^2x}-2\cos x-\dfrac{1}{\sin^2x}}{(2\tan x-2\sin x+\cot x)^2}$ $=\dfrac{2\sin^2x-2\cos x.\sin^2x\cos^2x-\cos^2x}{\sin^2x\cos^2x(2\tan x-2\sin x+\cot x)^2}$ $=\dfrac{2\sin^2x-\sin2x.\cos^2x.\sin x-\cos^2x}{\sin^2x\cos^2x(2\tan x-2\sin x+\cot x)^2}$ Bình luận
$y’=\dfrac{(2\tan x-2\sin x+\cot x)’}{(2\tan x-2\sin x+\cot x)^2}$
$=\dfrac{2(\tan x)’-2(\sin x)’+(\cot x)’}{(2\tan x-2\sin x+\cot x)^2}$
$=\dfrac{\dfrac{2}{\cos^2x}-2\cos x-\dfrac{1}{\sin^2x}}{(2\tan x-2\sin x+\cot x)^2}$
$=\dfrac{2\sin^2x-2\cos x.\sin^2x\cos^2x-\cos^2x}{\sin^2x\cos^2x(2\tan x-2\sin x+\cot x)^2}$
$=\dfrac{2\sin^2x-\sin2x.\cos^2x.\sin x-\cos^2x}{\sin^2x\cos^2x(2\tan x-2\sin x+\cot x)^2}$